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Home , Buffer solution, Weak base

5.111 Lecture Summary #22 Wednesday, October 31, 2014

Reading for Today: Sections 11.13, 11.18-11.19, 12.1-12.3 in 5th ed. (10.13, 10.18-10.19, 11.1-11.3 in 4th ed.) Reading for Lecture #23: Sections 12.4-12.6 in 5th ed. (4th ed: 11.4-11.6)

Topics: I. pH of salt II. Buffers!

I. pH of salt solutions

A salt is formed by the neutralization of an by a .

HCl + NaOH gives NaCl and H2O

The pH of salt in water is not always .

Salts that contain the conjugate of weak bases produce acidic aqueous solutions; so do salts that contain small, highly charged metal cations (e.g. Fe3+).

(Note: all Group 1 and 2 metals (e.g. Li+, Ca+2) and all metal cations with charge +1 (e.g. Ag+1) are neutral.)

Salts that contain the conjugate bases of weak acids produce basic aqueous .

1) NH4Cl (aq) will produce a(n) solution.

+ + NH4 Is NH4 a of a and -10 therefore a weak acid? Ka = 5.6 x 10

-5 Is NH3 a weak base? Kb = 1.8 x 10

Cl- Is Cl- a conjugate base of a weak acid and therefore a weak base?

7 Is HCl a weak acid? Ka = 10

-5 2) NaCH3COO (aq) will produce a(n) solution. Ka of CH3COOH is 1.76 x 10

Na+ Is Na+ a conjugate acid of a weak base and therefore acidic?

- - CH3COO Is CH3COO a conjugate base of a weak acid and therefore basic? Is CH3COOH is weak acid?

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3) General rule for compound XY

X+ Is X+ a conjugate acid of a weak base? If yes, then acidic; If no, neutral Y- Is Y- a conjugate base of a weak acid? If yes, then basic; If no, neutral

Overall: acidic+neutral=acidic; basic+neutral=basic;neutral+neutral=neutral

II. BUFFERS!

A is any solution that maintains an approximately pH despite small additions of acid and base.

An acid buffer: consists of a and its supplied as a salt. It buffers on the acidic side of neutral.

A base buffer: consists of a weak base and its conjugate acid supplied as a salt. It buffers on the basic side of neutral.

Acid Buffer Example: Mix with an acetate salt and get :

+ - CH3COOH (aq) + H2O (l) H3O (aq) + CH3CO2 (aq)

What happens if strong acid is added to a solution containing approximately equal amounts - of CH3CO2 and CH3COOH? • reaction + • The added H3O ions are effectively and the pH stays constant.

What happens if OH- base is added? - • The base removes a proton from CH3COOH to form H2O and CH3CO2 molecules. • The added ions are effectively removed and the pH stays constant.

Acid buffer action: The weak acid, HA, transfers protons to OH- ions supplied by strong - + base. The conjugate base, A , of the weak acid accepts protons from the H3O ions supplied by a strong acid.

A strong acid and the salt of its conjugate base don't make a good buffer. Why?

+ - Base Buffer Example: NH3 (aq) + H2O (l) NH4 (aq) + OH (aq)

+ When strong acid is added, NH3 accepts protons from incoming acid to make NH4 . When + strong base is added, NH4 donates a proton to form NH3 and H2O. pH remains the .

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Base buffer action: The weak base, B, accepts protons supplied by strong acid. The conjugate acid, BH+, of the weak base transfers protons to the OH- ions supplied by a strong base.

A buffer is a of weak conjugate acids and bases that stabilize the pH of a solution by providing a or for protons.

Buffers are important in biology! is buffered in the range of 7.35-7.45. Buffering agents: - H2CO3/HCO3

Sample Buffer Problem: Suppose 1.00 mol of HCOOH and 0.500 mol of NaHCOO are added -4 to water and diluted to 1.0 L. Calculate the pH. (Ka = 1.77 x 10 )

+ - HCOOH + H2O H3O + HCOO initial molarity 1.00 0 0.500 change in molarity -x +x +x equilibrium molarity

-4 Ka = 1.77 x 10 =

Using approximation that x is small compared to 1.00 and 0.500,

-4 Ka = 1.77 x 10 =

x= 3.54 x 10-4 M

Check assumption

3.54 x 10-4 =

pH =

Now –Calculate the pH given that 0.100 mol of a strong acid (HCl) had been included in the 1.0 L solution.

Because 0.100 mol of HCl reacts with number of moles of HCOO- to form moles of HCOOH:

For HCOO-, 0.500 mol - 0.100 mol = mol [HCOO-] = 0.400 mol/1.0 L =0.400 M

For HCOOH, 1.00 mol + 0.100 mol = mol [HCOOH] = 1.10 mol/1.0 L =1.10 M 3 + - HCOOH + H2O H3O + HCOO initial molarity change in molarity equilibrium molarity

-4 Ka = 1.77 x 10 =

Using approximation that x is small,

-4 -4 Ka = 1.77 x 10 = x= 4.87 x 10

Check assumption (5% rule)

+ pH = -log [H3O ] = 3.31

So addition of 0.10 mol of strong acid only changed pH from 3.45 to 3.31!

Designing a Buffer

- One must consider the relationship between the ratio of [HA] to [A ], pKa, and pH in designing a buffer. [H O+][A-] + - 3 HA (aq) + H2O <---> H3O (aq) + A (aq) Ka = [H O+][A-] + - K = 3 HA (aq) + H2O <---> H3O (aq) + A (aq) a [HA] [HA]

+ Rearrange: [H O ] = Ka x [HA] 3 + Rearrange: [H3O ] = Ka x [HA] [A-] - [A ] [HA] + Take logarithms of both sides: log [H3O ] = log Ka + log [HA] + - Take logarithms of both sides: log [H3O ] = logKa + log [A ] [A-]

+ [HA] Multiply by (-) : -log [H3O ] = -log Ka - log [HA] + - Multiply by (-) : -log[H3O ] = -log Ka - log [A ] [A-] [HA] That is: pH = pKa - log [HA]

That is: pH = pKa - log [A-] [A-] eq eq The values of [HA] and [A-] in the equation are at equilibrium.

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However, a weak acid HA typically loses only a tiny fraction of its protons.

Likewise, a weak base A- typically only accepts a tiny fraction of protons.

So initial is approximately __ to equilibrium concentration

~ [HA]0 So pH = pKa - log Henderson-Hasselbalch Equation - [A ]0

initial

+ - This assumption is valid when [H3O ] is compared to [HA] and [A ] (i.e. less than 5%).

Example: Design a buffer system with pH 4.60.

Acetic acid is suitable with a pKa of 4.75

A buffer solution is most effective in the range of pKa ±1 [CH3COOH]0 pH = pKa - log - [CH3COO ]0

[CH COOH] log 3 0 = pK - pH = 4.75 - 4.60 = 0.15 - a [CH3COO ]0

[CH COOH] 3 0 0.15 - = 10 = 1.4 [CH3COO ]0

The ratio is more important than the amounts used.

However, the amounts do affect the capacity of the buffer to changes in pH. Higher = more resistance to change.

If you use too low concentrations, the Henderson-Hasselbalch equation won’t be valid.

+ -5 For pH 4.60, [H3O ] is 2.5 x 10 .

-5 2.5 x 10 x 100% < 5% -4 need concentration > 5.0 x 10 M [HA] or [A-]

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5.111 Principles of Chemical Science Fall 2014

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